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A potentiometer wire of length 1.0 m has a resistance of 15 ohm. It is connected to a 5 V source in series with a resistance of 5 Omega. Determine the emf of the primary cell which has a balance point at 60 cm. |
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Answer» Solution :Current `I=V/(R_1+R_2)` `=5/(15+5)`=0.05 A Potential drop across the potentiometer wire V = IR= 0.25 × 15 = 3.75 V Potential gradient, `k=V/t=3.75/1.0 =3.75` V/m `therefore`Unknown emf of the CELL = KI = 3.75 × 0.6 = 2.25 V. |
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